A) \[2\pi \]
B) \[\pi \]
C) \[\frac{\pi }{4}\]
D) 0
Correct Answer: B
Solution :
Let \[I=\int_{0}^{\pi }{xf(\sin x)dx}=A\int_{0}^{\pi /2}{f(\sin x)dx}\] Now, \[2I=\int_{0}^{\pi }{xf(\sin x)dx+\int_{0}^{\pi }{(\pi -x)f[\sin (\pi -x)]dx}}\] \[=\int_{0}^{\pi }{\pi f(\sin x)dx}=\pi \int_{0}^{\pi }{f(\sin x)dx}\] Þ \[2I=2\pi \int_{0}^{\pi /2}{f(\sin x)dx}\] \ \[I=\pi \int_{0}^{\pi /2}{f(\sin x)dx}\]\[=A\int_{0}^{\pi }{f(\sin x)dx}\]. Hence \[A=\pi \].
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